# a compact metric space is second countable

###### Proposition.

###### Proof.

Let $(X,d)$ be a compact^{} metric space, and for each $n\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{+}$ define ${\mathrm{\pi \x9d\x92\x9c}}_{n}=\{B\beta \x81\u2019(x,1/n):x\beta \x88\x88X\}$, where $B\beta \x81\u2019(x,1/n)$ denotes the open ball centered about $x$ of http://planetmath.org/node/1296radius $1/n$. Each such collection^{} is an open cover of the compact space $X$, so for each $n\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{+}$ there exists a finite collection ${\mathrm{\beta \x84\neg}}_{n}\beta \x8a\x86{\mathrm{\pi \x9d\x92\x9c}}_{n}$ that $X$. Put $\mathrm{\beta \x84\neg}={\beta \x8b\x83}_{n=1}^{\mathrm{\beta \x88\x9e}}{\mathrm{\beta \x84\neg}}_{n}$. Being a countable^{} union of finite sets^{}, it follows that $\mathrm{\beta \x84\neg}$ is countable; we assert that it forms a basis for the metric topology on $X$. The first property of a basis is satisfied trivially, as each set ${\mathrm{\beta \x84\neg}}_{n}$ is an open cover of $X$. For the second property, let $x,{x}_{1},{x}_{2}\beta \x88\x88X$, ${n}_{1},{n}_{2}\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{+}$, and suppose $x\beta \x88\x88B\beta \x81\u2019({x}_{1},1/{n}_{1})\beta \x88\copyright B\beta \x81\u2019({x}_{2},1/{n}_{2})$. Because the sets $B\beta \x81\u2019({x}_{1},1/{n}_{1})$ and $B\beta \x81\u2019({x}_{2},1/{n}_{2})$ are open in the metric topology on $X$, their intersection^{} is also open, so there exists $\mathrm{{\rm O}\u0385}>0$ such that $B\beta \x81\u2019(x,\mathrm{{\rm O}\u0385})\beta \x8a\x86B\beta \x81\u2019({x}_{1},1/{n}_{1})\beta \x88\copyright B\beta \x81\u2019({x}_{2},1/{n}_{2})$. Select $N\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{+}$ such that $$. There must exist ${x}_{3}\beta \x88\x88X$ such that $x\beta \x88\x88B\beta \x81\u2019({x}_{3},1/2\beta \x81\u2019N)$ (since ${\mathrm{\beta \x84\neg}}_{2\beta \x81\u2019N}$ is an open cover of $X$). To see that $B\beta \x81\u2019({x}_{3},1/2\beta \x81\u2019N)\beta \x8a\x86B\beta \x81\u2019({x}_{1},1/{n}_{1})\beta \x88\copyright B\beta \x81\u2019({x}_{2},1/{n}_{2})$, let $y\beta \x88\x88B\beta \x81\u2019({x}_{3},1/2\beta \x81\u2019N)$. Then we have

$$ | (1) |

so that $y\beta \x88\x88B\beta \x81\u2019(x,\mathrm{{\rm O}\u0385})$, from which it follows that $y\beta \x88\x88B\beta \x81\u2019({x}_{1},1/{n}_{1})\beta \x88\copyright B\beta \x81\u2019({x}_{2},1/{n}_{2})$, hence that $B\beta \x81\u2019({x}_{3},1/2\beta \x81\u2019N)\beta \x8a\x86B\beta \x81\u2019({x}_{1},1/{n}_{1})\beta \x88\copyright B\beta \x81\u2019({x}_{2},1/{n}_{2})$. Thus the countable collection $\mathrm{\beta \x84\neg}$ forms a basis for a topology^{} on $X$; the verification that the topology by $\mathrm{\beta \x84\neg}$ is in fact the metric topology follows by an to that used to verify the second property of a basis, and completes^{} the proof that $X$ is second countable.
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It is worth nothing that, because a countable union of countable sets is countable, it would have been sufficient to assume that $(X,d)$ was a LindelΓΆf space.

Title | a compact metric space is second countable |

Canonical name | ACompactMetricSpaceIsSecondCountable |

Date of creation | 2013-03-22 17:00:49 |

Last modified on | 2013-03-22 17:00:49 |

Owner | azdbacks4234 (14155) |

Last modified by | azdbacks4234 (14155) |

Numerical id | 17 |

Author | azdbacks4234 (14155) |

Entry type | Theorem |

Classification | msc 54D70 |

Related topic | MetricSpace |

Related topic | Compact |

Related topic | Lindelof |

Related topic | Ball |

Related topic | basisTopologicalSpace |

Related topic | Cover |

Related topic | BasisTopologicalSpace |